Problem

for $\Phi(\cdot)$ : standard norm CDF. Unfortunately there is no spoon-feeding solution for this problem in the Wiki page for the List of Integrals of Gaussian functions. For those who may flag this post, this post does not seem to address my question due to some misunderstandings on OP's part.

Try

But I tried with the indefinite integral of $\int \Phi(ax+b)dx$, which is given by

$\begingroup$FWIW, there is some confusion in this post concerning the roles of $a,b,c,d.$ I don't think that makes the question incomprehensible, but readers should be careful.$\endgroup$
– whuber♦Jan 25 at 16:31

In the question with positive $a$ and $c$ and $F=\Phi,$ we have $E[X] = 0,$ reducing the integral to $b/a - d/c.$ When $a=c$ this simplifies to $(b-d)/a,$ exactly as suggested in the question.

This result isn't quite the most general one: when $a=c,$ the result holds in the form $(b-d)/a$ even when $X$ does not have a finite expectation. This is most easily seen by integrating the quantile function $F^{-1}:$ see https://stats.stackexchange.com/a/18439/919.

In case any of these manipulations appear doubtful, here is numerical confirmation using a host of different distributions (some, like the Pareto and Student t, have infinite variance; others--the versions of a Binomial and Poisson distribution--are discrete). Each "Example ..." column corresponds to these randomly-chosen $(a,b,c,d):$

Each pair of lines shows the integral's value followed by the formula's value; they agree in every case.

Here is the code that performed these computations. Notice how the expectation $E[X]$ is carried out with the integral $(*)$ in the function g. The blind integration of the discrete distribution functions in g can be a little delicate; this is handled by increasing the default number of subdivisions from 100 to 1000, but could be further improved by using a finite lower limit of integration (thereby giving the routine a decent hint concerning the scale of the calculation).